如何将来自javascript函数的值输入到另一个php页面以将其保存到DB中

时间:2014-07-23 05:14:20

标签: javascript php jquery html ajax

这是我在单击添加按钮时在表上添加动态行的代码。现在我需要将输入的值输入文本框并提交给DB。

<div id="addinput">
    <p>
    <button name="add" onclick="myFunction(this)" value="'.$btn_count.'"><font color="#FF0000" size="4">+</font></button>
    </p>
    </div>
    </td>

这是我向桌面添加动态行的功能,就在我们点击

的行的下方
<script>
function myFunction(y)
{

var f = y.parentNode;
var d = f.parentNode;
var g = d.parentNode;
var c =g.parentNode.rowIndex;   


var table = document.getElementById("myTable");

var row = table.insertRow(c+1);


var cell1 = row.insertCell(0);
var cell2 = row.insertCell(1);
var cell3 = row.insertCell(2);
var cell4 = row.insertCell(3);
var cell5 = row.insertCell(4);
var cell6 = row.insertCell(5);
var cell7 = row.insertCell(6);

cell1.innerHTML = "<form name='frm_sub' id='frm_id' action='#'><input type=text  name='txt_id' id='id_id' size=10 placeholder=Id>";
cell2.innerHTML = "<input type=text name='txt_item' id='id_item' size=10 placeholder=Item>";
cell3.innerHTML = "<input type=text name='txt_cost' id='id_cost' size=10 placeholder=0.00>";
cell4.innerHTML = "<input type=text name='txt_wp' id='id_wp' size=10 placeholder=0.00>";
cell5.innerHTML = "<input type=text name='txt_rp' id='id_rp' size=10 placeholder=0.00>";

cell6.innerHTML = "<input type='button' name='submit' id='id_submitBtn' value='Save' size=10 onclick='form_submit()' ></form>";


document.getElementById("myTable");

}

function form_submit()
{


var item_name = $("input#id_item").val();

var cost = $("input#id_cost").val();

var wp = $("input#id_wp").val();

var rp = $("input#id_rp").val();

var dataString = 'item_name='+ item_name + '&cost=' + cost + '&wp=' + wp+ '&rp=' + rp;

alert(dataString);



$.ajax({

        type: "POST",
        url:newAdd.php,
        dataType: "json",
        type: "POST",
        data: {

            items : $("#id_item").val(),           
            cost : $("#id_cost").val(),
            wp : $("#id_wp").val(),
            rp : $("#id_rp").val(),
        },
        success: function () {
            alert("Success");
        }
    });



}

在newAdd.php我写了

echo$itemName =$_POST["txt_item"];
$insert = mysql_query('INSERT INTO item (item) VALUES ('$itemName');

但这不起作用..

2 个答案:

答案 0 :(得分:4)

if you are getting that alert  then this will work 


    function form_submit()
    {


    var item_name = $("input#id_item").val();

    var cost = $("input#id_cost").val();

    var wp = $("input#id_wp").val();

    var rp = $("input#id_rp").val();

    var dataString = 'item_name='+ item_name + '&cost=' + cost + '&wp=' + wp+ '&rp=' + rp;
    $.ajax({
        type: "post",
        url: "newAdd.php",
        data:dataString,
        success: function(msg){ 
        alert('success');   

        }
        });



    }

答案 1 :(得分:0)

试试这个ajax请求。希望这会对你有所帮助 -

$.ajax({
        type: "POST",
        url:'newAdd.php',
        dataType: "json",
        data: {
            items : $("#id_item").val(),           
            cost : $("#id_cost").val(),
            wp : $("#id_wp").val(),
            rp : $("#id_rp").val()
        },
        success: function () {
            alert("Success");
        }
    });